# Matrix ODE

by Manchot
Tags: matrix
 P: 728 I'm trying to find a general solution for the logistic ODE $\frac{dU}{dx}=A(I-U)U$, where A and U are square matrices and x is a scalar parameter. Inspired by the scalar equivalent I guessed that $U=(I+e^{-Ax})^{-1}$ is a valid solution; however, $U=(I+e^{-Ax+B})^{-1}$ is not when U and A don't commute. Any ideas?
 PF Gold P: 354 The general solution to the scaler equation is: E^(A x)/(E^(A x) + E^C) where C is a constant. Maybe this can lead to a similar solution for the matricial version?
 PF Gold P: 354 If U is a function of A, then U commutes with A.
 P: 728 Matrix ODE I tried all sorts of versions of the scalar equation, maajdl. They all run into the same commutation problem. Unfortunately, U is a function of both A and the initial condition, which means that it doesn't commute with A unless the initial condition does.
 PF Gold P: 354 Could that help? Assuming: A = M-1DM where D is a diagonal matrix V = MUM-1 The ODE becomes: dV/dx = D(I-V)V
 P: 728 Yeah, I tried diagonalizing both A and the initial condition. No dice.
 PF Gold P: 354 What is your practical goal? Why do you need a formal solution?

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